Question

Question 4 The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:

Length (in mm) Number of leaves
118 − 126 3
127 − 135 5
136 − 144 9
145 − 153 12
154 − 162 5
163 − 171 4
172 − 180 2

(i) Draw a histogram to represent the given data.
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

Class 9 - Math - Statistics Page 259"

Answer 1

The given data is in discontinuous form.
Now, 1/2 = 0.5Add 0.5 to upper limit  and subtract it from lower class limit.

Now the table looks like this -->
117.5 - 126.5 ---3
126.5 - 135.5 ---5
135.5 - 144.5 ---9
144.5 - 153.5 ---12
153.5 - 162.5 --- 5
162.5 - 171.5 ---41
71.5 - 180.5 ---2

ii) This graph can also be drawn using frequency polygon.

iii) No its not correct. We can say that maximum number of leaves have their length between the range of 144.5 - 153.5 only.

Hope This Helps You!

Answer 2

Histogram 

Histogram is a graphical representation of a grouped frequency distribution in exclusive form ( 0-10, 10-20) with continuous classes in the form of rectangles with class intervals as bases and the corresponding frequencies as Heights.

There is no gap between any two consecutive rectangles.

 

In histogram we draw a kink.

Kink i.e.  break is indicated near the origin to signify that a graph is drawn not at the origin.

If the first class interval is not starting from zero , then we show it on the graph by making a kink or a break on the axis.

frequency polygon

A frequency polygon is the polygon obtained by joining the midpoints of upper horizontal sides of all the rectangles in the histogram. Frequency polygons can also be drawn independently without drawing histogram.

 

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 Solution:

 

We know that the areas of the rectangles are proportional to the frequencies in a histogram .

Here, the given frequency distribution is not continuous as it is in inclusive form(0-10, 11-20).

To Convert the intervals in continuous form firstly we find  the difference between upper limit of a class and the lower limit of its succeeding class then add and subtract half of this difference to upper limit and lower limit.

 

For continuous distribution we get first interval as

(118-0.5) – (126+0.5)= 117.5 -126.5.

The class width in this case is 9.

 

So we get the following modified table of given data in the attachment.

 

Now we can draw histogram for given data (in the attachment)

 

ii)

Yes other suitable graphical representation for the same data is frequency polygon.

iii)

No, Because the maximum number of leaves have their lengths lying in the interval 145 -153.

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Hope this will help you.....